Calculate the vapor pressure of a solution containing 24.6 g of glycerin (C3H8O3) in 134 mL of water at 30.0 ∘C. The vapor pressure of pure water at this temperature is 31.8 torr. Assume that glycerin is not volatile and dissolves molecularly (i.e., it is not ionic) and use a density of 1.00 g/mL for the water.

Respuesta :

Answer : The vapor pressure of a solution is, 30.687 torr

Explanation :

First we have to calculate the moles of glycerin.

[tex]\text{Moles of }C_3H_8O_3=\frac{\text{Mass of }C_3H_8O_3}{\text{Molar mass of }C_3H_8O_3}=\frac{24.6g}{92.09g/mole}=0.267moles[/tex]

Now we have to calculate the mass of water.

[tex]\text{Mass of }H_2O=\text{Density of }H_2O\times \text{Volume of }H_2O[/tex]

[tex]\text{Mass of }H_2O=(1.00g/ml)\times (134ml)=134g[/tex]

Now we have to calculate the moles of water.

[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{134g}{18g/mole}=7.444moles[/tex]

Now we have to calculate the mole fraction of water.

[tex]\text{Mole fraction of water}=\frac{\text{Moles of water}}{\text{Moles of water}+\text{Moles of glycerin}}[/tex]

[tex]\text{Mole fraction of water}=\frac{7.444mole}{7.444mole+0.267mole}=0.965[/tex]

Now we have to calculate the vapor pressure of the solution.

According to the Raoult's law,

[tex]p_A=X_A\times p^o_A[/tex]

where,

[tex]p_A[/tex] = vapor pressure of solution = ?

[tex]p^o_A[/tex] = vapor pressure of pure water= 31.8 torr

[tex]X_A[/tex] = mole fraction of water = 0.965

Now put all the given values in this formula, we get the vapor pressure of solution.

[tex]p_A=0.965\times 31.8\text{ torr}[/tex]

[tex]p_A=30.687\text{ torr}[/tex]

Therefore, the vapor pressure of a solution is, 30.687 torr

Considering the Raoult's Law, the vapor pressure of a solution that containing 24.6 g of glycerin in 134 mL of water at 30.0 ∘C is 30.687 torr.

Raoult's Law

Raoult's Law is a gas law that relates the vapor pressure and mole fraction of each gas in a solution (solution).

If a solute has a measurable vapor pressure, the vapor pressure of its solution is always less than that of the pure solvent. Thus, the relationship between the vapor pressure of the solution and the vapor pressure of the solvent depends on the concentration of the solute in the solution.

Raoult's law allows us to calculate the vapor pressure of a substance when it is part of an ideal solution, knowing its vapor pressure when it is pure (at the same temperature) and the composition of the ideal solution in terms of molar fraction.

Then, this law establishes as a conclusion that: “In an ideal solution, the partial pressures of each component in the vapor are directly proportional to their respective mole fractions in the solution”.

Mathematically, Raoult's law states that the partial pressure of a solvent above a solution P₁ is given by the vapor pressure of the pure solvent P₁⁰ multiplied by the mole fraction of the solvent in the solution X₁:

P₁= P₁⁰ × X₁

Vapor pressure in this case

The mole fraction is defined as the ratio of moles of solute to total moles of solution.

So, the moles of glycerin (solute) is calculated as follow:

[tex]moles of glycerin=\frac{mass of glycerin}{molar mass of glycerin}=\frac{24.6 g}{92.09 \frac{g}{mole} } =[/tex] 0.267 moles

Being the density the measure of the amount of mass in a certain volume of a substance, the mass of water (solvent) in the solution is calculated as:

mass of water= volume of water× density of water

mass of water= 134 mL× 1 [tex]\frac{g}{mL}[/tex]= 134 g

So, the moles of water (solvent) is calculated as follow:

[tex]moles of water=\frac{mass of water}{molar mass of water}=\frac{134 g}{18 \frac{g}{mole} } =[/tex] 7.44 moles

Then, the mole  fraction of water can be calculated as:

[tex]Mole fraction of water=\frac{moles of water}{moles of glycerin + moles of water}[/tex]

[tex]Mole fraction of water=\frac{7.44 moles}{0.267 moles + 7.44 moles}[/tex]

Solving:

mole fraction of water= 0.965

Then, you know that:

  • [tex]P_{water} =[/tex] ?
  • [tex]P_{water} ^{0}=[/tex] 31.8 torr
  • [tex]x_{water}=[/tex] 0.965

Replacing in Raoult's Law:

[tex]P_{water} =[/tex] 31.8 torr × 0965

Solving:

[tex]P_{water} =[/tex] 30.687 torr

In summary, the vapor pressure of a solution that containing 24.6 g of glycerin in 134 mL of water at 30.0 ∘C is 30.687 torr.

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